1. Reaction during electrolysis of aqueous NaCl

In aqueous NaCl, at the cathode water is reduced: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

Thus, 1 mole of electron produces 1 mole of $OH^-$.

2. Use the final pH to find $[OH^-]$

Given final pH = 12, $pOH = 14 - 12 = 2$ $[OH^-] = 10^{-2}\,\text{M}$

Volume of solution: $V = 600\,\text{mL} = 0.6\,\text{L}$

So moles of $OH^-$ formed are: $n(OH^-) = [OH^-] \times V = 10^{-2} \times 0.6 = 6\times 10^{-3}\,\text{mol}$

3. Relate moles of $OH^-$ to charge passed

Since 1 mole $e^-$ gives 1 mole $OH^-$, $n(e^-) = 6\times 10^{-3}\,\text{mol}$

Charge passed: $Q = nF = 6\times 10^{-3} \times 96500 \approx 579\,\text{C}$

4. Find current

Time of electrolysis: $t = 5\,\text{min} = 300\,\text{s}$

Using $Q = It$, $I = \frac{Q}{t} = \frac{579}{300} \approx 1.93\,\text{A}$

Nearest integer: $\boxed{2}$

5. Comparison with stored answer

Derived answer = $2$

Stored correct answer = $2$

Hence, the derived answer agrees with the stored correct answer.